Summary: A PROPERTY OF STRICTLY SINGULAR 1-1 OPERATORS
GEORGE ANDROULAKIS, PER ENFLO
Abstract We prove that if T is a strictly singular 1-1 operator defined on an infinite di-
mensional Banach space X, then for every infinite dimensional subspace Y of X there exists
an infinite dimensional subspace Z of X such that Z Y is infinite dimensional, Z contains
orbits of T of every finite length and the restriction of T on Z is a compact operator.
1. Introduction
An operator on an infinite dimensional Banach space is called strictly singular if it fails to
be an isomorphism when it is restricted to any infinite dimensional subspace (by "operator"
we will always mean a "continuous linear map"). It is easy to see that an operator T on
an infinite dimensional Banach space X is strictly singular if and only if for every infinite
dimensional subspace Y of X there exists an infinite dimensional subspace Z of Y such
that the restriction of T on Z, T|Z : Z X, is a compact operator. Moreover, Z can
be assumed to have a basis. Compact operators are special examples of strictly singular
operators. If 1 p < q then the inclusion map ip,q : p q is a strictly singular
(non-compact) operator. A Hereditarily Indecomposable (H.I.) Banach space is an infinite
dimensional space such that no subspace can be written as a topological sum of two infinite
dimensional subspaces. W.T. Gowers and B. Maurey constructed the first example of an H.I.
space [8]. It is also proved in [8] that every operator on a complex H.I. space can be written
as a strictly singular perturbation of a multiple of the identity. If X is a complex H.I. space